1.次の定積分を求めなさい。
(1)\(\displaystyle \int_2^4 x^3dx\)
\(\displaystyle =\left[\frac{1}{4}x^4\right]_2^4\)
\(\displaystyle =\frac{1}{4}・4^4-\frac{1}{4}・2^4\)
\(=60\)
(2)\(\displaystyle \int_1^2 \frac{1}{\sqrt{x}}dx\)
\(\displaystyle =[2x^{\frac{1}{2}}]_1^2\)
\(\displaystyle =2・2^{\frac{1}{2}}-2・1^{\frac{1}{2}}\)
\(=2\sqrt{2}-2\)
(3)\(\displaystyle \int_0^{\frac{\pi}{4}} \frac{1}{\cos^2x}dx\)
\(\displaystyle =[\tan x]_0^{\frac{\pi}{4}}\)
\(=1-0\)
\(=1\)
(4)\(\displaystyle \int_{-2}^{0} e^xdx\)
\(\displaystyle =[e^x]_{-2}^{0}\)
\(\displaystyle =e^0-e^{-2}\)
\(\displaystyle =1-\frac{1}{e^2}\)
(5)\(\displaystyle \int_1^{2} \frac{dx}{x^2}\)
\(\displaystyle =-[x^{-1}]_1^{2}\)
\(\displaystyle =-\left(\frac{1}{2}-1\right)\)
\(\displaystyle =\frac{1}{2}\)
(6)\(\displaystyle \int_1^{8} \sqrt[3]{x}dx\)
\(\displaystyle =\frac{3}{4}[x^{\frac{4}{3}}]_1^{8}\)
\(\displaystyle =\frac{3}{4}(16-1)\)
\(\displaystyle =\frac{45}{4}\)
(7)\(\displaystyle \int_0^{\frac{\pi}{2}} \cos xdx\)
\(\displaystyle =[\sin x]_0^{\frac{\pi}{2}}\)
\(=1-0\)
\(=1\)
(8)\(\displaystyle \int_0^{1} e^xdx\)
\(\displaystyle =[e^x]_0^{1}\)
\(=e^1-e^0\)
\(=e-1\)
(9)\(\displaystyle \int_{-2}^{-1}\frac{dx}{x}\)
\(\displaystyle =[\log|x|]_{-2}^{-1}\)
\(=\log1-\log2\)
\(=-\log2\)
(10)\(\displaystyle \int_{-1}^{1}2^xdx\)
\(\displaystyle =\left[\frac{2^x}{\log2}\right]_{-1}^{1}\)
\(\displaystyle =\frac{2}{\log2}-\frac{1}{2\log2}\)
\(\displaystyle =\frac{3}{2\log2}\)
(11)\(\displaystyle \int_1^{2} \sqrt{x+1}dx\)
\(\displaystyle =\frac{2}{3}[(x+1)^{\frac{3}{2}}]_1^{2}\)
\(\displaystyle =\frac{2}{3}(3^{\frac{3}{2}}-2^{\frac{3}{2}})\)
\(\displaystyle =\frac{2}{3}(3\sqrt{3}-2\sqrt{2})\)
(12)\(\displaystyle \int_0^{1} (2x+1)^3dx\)
\(\displaystyle =\frac{1}{4}・\frac{1}{2}[(2x+1)^4]_0^{1}\)
\(\displaystyle =\frac{1}{8}(81-1)\)
\(=10\)
(13)\(\displaystyle \int_{-1}^{1} (e^x-e^{-x})dx\)
\(\displaystyle =[(e^x+e^{-x})]_{-1}^{1}\)
\(\displaystyle =\left(e+\frac{1}{e}\right)-\left(\frac{1}{e}+e\right)\)
\(=0\)
(14)\(\displaystyle \int_0^{\pi} \sin2xdx\)
\(\displaystyle =\frac{1}{2}[-\cos2x]_0^{\pi}\)
\(\displaystyle =\frac{1}{2}(-1+1)\)
\(=0\)
(15)\(\displaystyle \int_2^{3} \frac{x+3}{x^2-1}dx\)
\(\displaystyle =\int_2^{3} \left(\frac{2}{x-1}-\frac{1}{x+1}\right)dx\)
\(\displaystyle =[2\log|x-1|]_2^{3}-[\log|x+1|]_2^{3}\)
\(=2\log2-0-(\log4-\log3)\)
\(=\log3\)
(16)\(\displaystyle \int_0^{2\pi} \cos^2xdx\)
\(\displaystyle =\frac{1}{2}\int_0^{2\pi} (\cos2x+1)dx\)
\(\displaystyle =\frac{1}{2}\left[\frac{1}{2}\sin2x+x\right]_0^{2\pi}\)
\(\displaystyle =\frac{1}{2}・2\pi\)
\(=\pi\)
(17)\(\displaystyle \int_0^{\frac{\pi}{4}} \sin4x\cos3xdx\)
\(\displaystyle =\frac{1}{2}\int_0^{\frac{\pi}{4}} (\sin7x+\sin x)dx\)
\(\displaystyle =\frac{1}{2}\left[-\frac{1}{7}\cos7x\right]_0^{\frac{\pi}{4}}+\frac{1}{2}[-\cos x]_0^{\frac{\pi}{4}}\)
\(\displaystyle =-\frac{1}{14}\left(\frac{1}{\sqrt{2}}-1\right)-\frac{1}{2}\left(\frac{1}{\sqrt{2}}-1\right)\)
\(\displaystyle =\frac{4-2\sqrt{2}}{7}\)
(18)\(\displaystyle \int_0^{\frac{\pi}{2}} \sin4x\cos2xdx\)
\(\displaystyle =\frac{1}{2}\int_0^{\frac{\pi}{2}} (\sin6x+\sin2x)dx\)
\(\displaystyle =\frac{1}{2}\left[-\frac{1}{6}\cos6x\right]_0^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{1}{2}\cos2x\right]_0^{\frac{\pi}{2}}\)
\(\displaystyle =\frac{1}{2}\left(\frac{1}{6}+\frac{1}{6}\right)-\frac{1}{2}\left(-\frac{1}{2}-\frac{1}{2}\right)\)
\(\displaystyle =\frac{2}{3}\)
(19)\(\displaystyle \int_{-1}^{2} \sqrt{|x|}dx\)
\(\displaystyle =\int_{-1}^{0} \sqrt{-x}dx+\int_0^{2} \sqrt{x}dx\)
\(\displaystyle =-\left[\frac{2}{3}(-x)^{\frac{3}{2}}\right]_{-1}^{0}+\left[\frac{2}{3}x^{\frac{3}{2}}\right]_0^{2}\)
\(\displaystyle =-0+\frac{2}{3}+\frac{4}{3}\sqrt{2}-0\)
\(\displaystyle =\frac{2+4\sqrt{2}}{3}\)
(20)\(\displaystyle \int_0^{4} |\sqrt{x}-1|dx\)
\(\displaystyle =-\int_0^{1}(\sqrt{x}-1)dx+\int_1^{4}(\sqrt{x}-1)dx\)
\(\displaystyle =-\left[\frac{2}{3}x^{\frac{3}{2}}-x\right]_0^{1}+\left[\frac{2}{3}x^{\frac{3}{2}}-x\right]_1^{4}\)
\(\displaystyle =-\left(\frac{2}{3}-1\right)+\left(\frac{16}{3}-4\right)-\left(\frac{2}{3}-1\right)\)
\(=2\)
(21)\(\displaystyle \int_{-1}^{2} |e^x-1|dx\)
\(\displaystyle =-\int_{-1}^{0}(e^x-1)dx+\int_0^{2}(e^x-1)dx\)
\(\displaystyle =-[e^x-x]_{-1}^{0}+[e^x-x]_0^{2}\)
\(=-(1-e^{-1}-1)+(e^2-2-1)\)
\(=e^2+\frac{1}{e}-3\)
